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د. بســام كحـالــه Dr. Bassam Kahhaleh

د. بســام كحـالــه Dr. Bassam Kahhaleh. 4241. Digital Logic Design. Chapter 1:. 4241 - Digital Logic Design. Binary Systems. Analog. Discrete. Digital Systems. Discrete Data Examples: 26 letters of the alphabet (A, B … etc) 10 decimal digits (0, 1, 2 … etc) Combine together

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د. بســام كحـالــه Dr. Bassam Kahhaleh

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  1. د. بســام كحـالــه Dr. Bassam Kahhaleh 4241 Digital Logic Design

  2. Chapter 1: 4241 - Digital Logic Design Binary Systems

  3. Analog Discrete Digital Systems • Discrete Data • Examples: • 26 letters of the alphabet (A, B … etc) • 10 decimal digits (0, 1, 2 … etc) • Combine together • Words are made of letters (University … etc) • Numbers are made of digits (4241 … etc) • Binary System • Only ‘0’ and ‘1’ digits • Can be easily implemented in electronic circuits

  4. 2 1 0 -1 -2 100 10 1 0.1 0.01 500 10 2 0.7 0.04 Decimal Number System • Base (also called radix) = 10 • 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } • Digit Position • Integer & fraction • Digit Weight • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 5 1 2 7 4 d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2 (512.74)10

  5. 64 8 1 1/8 1/64 2 1 0 -1 -2 Octal Number System • Base = 8 • 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 5 1 2 7 4 5 *82+1 *81+2 *80+7 *8-1+4 *8-2 =(330.9375)10 (512.74)8

  6. 4 2 1 1/2 1/4 2 1 0 -1 -2 Binary Number System • Base = 2 • 2 digits { 0, 1 }, called binary digits or “bits” • Weights • Weight = (Base) Position • Magnitude • Sum of “Bit x Weight” • Formal Notation • Groups of bits 4 bits = Nibble 8 bits = Byte 1 0 1 0 1 1 *22+0 *21+1 *20+0 *2-1+1 *2-2 =(5.25)10 (101.01)2 1 0 1 1 1 1 0 0 0 1 0 1

  7. 256 16 1 1/16 1/256 2 1 0 -1 -2 Hexadecimal Number System • Base = 16 • 16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 1 E 5 7 A 1 *162+14 *161+5 *160+7 *16-1+10 *16-2 =(485.4765625)10 (1E5.7A)16

  8. The Power of 2 Kilo Mega Giga Tera

  9. Addition • Decimal Addition 1 1 Carry 5 5 + 5 5 1 1 0 = Ten ≥Base  Subtract a Base

  10. Binary Addition • Column Addition 1 1 1 1 1 1 = 61 = 23 1 1 1 1 0 1 + 1 0 1 1 1 = 84 1 0 1 0 1 0 0 ≥ (2)10

  11. Binary Subtraction • Borrow a “Base” when needed 1 2 = (10)2 2 0 2 0 0 2 1 = 77 = 23 0 0 1 1 0 1 − 1 0 1 1 1 0 1 1 0 1 1 0 = 54

  12. Binary Multiplication • Bit by bit 1 0 1 1 1 x 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0

  13. Number Base Conversions Evaluate Magnitude Octal (Base 8) Evaluate Magnitude Decimal (Base 10) Binary (Base 2) Hexadecimal (Base 16) Evaluate Magnitude

  14. Decimal (Integer) to Binary Conversion • Divide the number by the ‘Base’ (=2) • Take the remainder (either 0 or 1) as a coefficient • Take the quotient and repeat the division Example: (13)10 Coefficient Quotient Remainder 13 / 2 =6 1a0 =1 6 / 2 =3 0a1 =0 3 / 2 =1 1a2 =1 1 / 2 =0 1a3 =1 Answer: (13)10= (a3 a2 a1 a0)2= (1101)2 MSB LSB

  15. Decimal (Fraction) to Binary Conversion • Multiply the number by the ‘Base’ (=2) • Take the integer (either 0 or 1) as a coefficient • Take the resultant fraction and repeat the division Example: (0.625)10 Coefficient Integer Fraction a-1 =1 0.625 * 2 =1 . 25 0.25 * 2 =0 . 5a-2 =0 0.5 * 2 =1 . 0a-3 =1 Answer: (0.625)10= (0.a-1 a-2 a-3)2= (0.101)2 MSB LSB

  16. Decimal to Octal Conversion Example: (175)10 Coefficient Quotient Remainder 175 / 8 =21 7a0 =7 21 / 8 =2 5a1 =5 2 / 8 =0 2a2 =2 Answer: (175)10= (a2 a1 a0)8= (257)8 Example: (0.3125)10 Coefficient Integer Fraction a-1 =2 0.3125 * 8 =2 . 5 0.5 * 8 =4 . 0a-2 =4 Answer: (0.3125)10= (0.a-1 a-2 a-3)8= (0.24)8

  17. Binary − Octal Conversion • 8 = 23 • Each group of 3 bits represents an octal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 2 6 . 2 )8 Works both ways (Binary to Octal & Octal to Binary)

  18. Binary − Hexadecimal Conversion • 16 = 24 • Each group of 4 bits represents a hexadecimal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 1 6 . 4 )16 Works both ways (Binary to Hex & Hex to Binary)

  19. Octal − Hexadecimal Conversion • Convert to Binary as an intermediate step Example: ( 2 6 . 2 )8 Assume Zeros Assume Zeros ( 01 0 1 1 0 . 0 1 0 )2 ( 1 6 . 4 )16 Works both ways (Octal to Hex & Hex to Octal)

  20. Decimal, Binary, Octal and Hexadecimal

  21. Complements • 1’s Complement (Diminished Radix Complement) • All ‘0’s become ‘1’s • All ‘1’s become ‘0’s Example (10110000)2  (01001111)2 If you add a number and its 1’s complement … 1 0 1 1 0 0 0 0 +0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1

  22. Complements • 2’s Complement (Radix Complement) • Take 1’s complement then add 1 • Toggle all bits to the left of the first ‘1’ from the right Example: Number: 1’s Comp.: OR 1 0 1 1 0 0 0 0 0 1 0 0 1 1 1 1 + 1 1 0 1 1 0 0 0 0 0101 0 0 0 0 0 1 0 1 0 0 0 0

  23. Negative Numbers • Computers Represent Information in ‘0’s and ‘1’s • ‘+’ and ‘−’ signs have to be represented in ‘0’s and ‘1’s • 3 Systems • Signed Magnitude • 1’s Complement • 2’s Complement All three use the left-most bit to represent the sign: • ‘0’  positive • ‘1’  negative

  24. Signed Magnitude Representation • Magnitude is magnitude, does not change with sign (+3)10 ( 0 0 1 1 )2 (−3)10  ( 1 0 1 1 )2 • Can’t include the sign bit in ‘Addition’ S Magnitude (Binary) Sign Magnitude 0 0 1 1  (+3)10 + 10 1 1  (−3)10 1 1 1 0  (−6)10

  25. Sign and Magnitude Representation High order bit is sign: 0 = positive (or zero), 1 = negative Three low order bits is the magnitude: 0 (000) thru 7 (111) Number range for n bits = +/-2n-1 -1 Two representations for 0

  26. 1’s Complement Representation • Positive numbers are represented in “Binary” • Negative numbers are represented in “1’s Comp.” (+3)10 (0 011)2 (−3)10  (1 100)2 • There are 2 representations for ‘0’ (+0)10  (0 000)2 (−0)10  (1 111)2 0 Magnitude (Binary) 1 Code (1’s Comp.)

  27. 1’s Complement Range • 4-Bit Representation 24 = 16 Combinations − 7 ≤ Number ≤ + 7 −23+1 ≤ Number ≤ +23 − 1 • n-Bit Representation −2n−1+1 ≤ Number ≤ +2n−1 − 1

  28. One’s Complement Representation • Subtraction implemented by addition & 1's complement • Still two representations of 0! This causes some problems • Some complexities in addition

  29. 2’s Complement Representation • Positive numbers are represented in “Binary” • Negative numbers are represented in “2’s Comp.” (+3)10 (0 011)2 (−3)10  (1 101)2 • There is 1 representation for ‘0’ (+0)10  (0 000)2 (−0)10  (0 000)2 0 Magnitude (Binary) 1 Code (2’s Comp.) 1’s Comp. 1 1 1 1 + 1 1 0 0 0 0

  30. 2’s Complement Range • 4-Bit Representation 24 = 16 Combinations − 8 ≤ Number ≤ + 7 −23≤ Number ≤ + 23 − 1 • n-Bit Representation −2n−1≤ Number ≤ + 2n−1 − 1

  31. Two’s Complement Representation • Only one representation for 0 • One more negative number than positive number like 1's comp except shifted one position clockwise

  32. 0 0 1 1 1 0 Number Representations • 4-Bit Example

  33. Binary Subtraction Using 1’s Comp. Addition • Change “Subtraction” to “Addition” • If “Carry” = 1then add it to theLSB, and the resultis positive(in Binary) • If “Carry” = 0then the resultis negative(in 1’s Comp.) (5)10 – (6)10 (5)10 – (1)10 (+5)10 + (-6)10 (+5)10 + (-1)10 0 1 0 1 0 1 0 1 + 1 0 0 1 + 1 1 1 0 0 1 1 1 0 0 0 1 1 1 + 1 1 1 0 0 1 0 0 + 4 − 1

  34. Binary Subtraction Using 2’s Comp. Addition • Change “Subtraction” to “Addition” • If “Carry” = 1ignore it, and the result is positive(in Binary) • If “Carry” = 0then the resultis negative(in 2’s Comp.) (5)10 – (6)10 (5)10 – (1)10 (+5)10 + (-6)10 (+5)10 + (-1)10 0 1 0 1 0 1 0 1 + 1 0 1 0 + 1 1 1 1 0 1 1 1 1 1 0 1 0 0 − 1 + 4

  35. Binary Codes • Group of n bits • Up to 2ncombinations • Each combination represents an element of information • Binary Coded Decimal (BCD) • Each Decimal Digit is represented by 4 bits • (0 – 9)  Valid combinations • (10 – 15)  Invalid combinations

  36. BCD Addition • One decimal digit + one decimal digit • If the result is 1 decimal digit ( ≤ 9 ), then it is a simple binary addition Example: • If the result is two decimal digits ( ≥ 10 ), then binary addition gives invalid combinations Example: 5 + 3 8 0 1 0 1 + 0 0 1 1 1 0 0 0 5 + 5 1 0 0 1 0 1 + 0 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0

  37. BCD Addition • If the binary resultis greater than 9,correct the result byadding 6 5 + 5 1 0 0 1 0 1 + 0 1 0 1 1 0 1 0 + 0 1 1 0 0 0 0 1 0 0 0 0 Multiple Decimal Digits Two Decimal Digits 3 5 1 0 0 1 1 0 1 0 1 0 0 0 1

  38. Gray Code • One bit changes fromone code to the nextcode • Different than Binary

  39. ASCII Code American Standard Code for Information Interchange

  40. Error Detecting Codes • Parity One bit added to a group of bits to make the total number of ‘1’s (including the parity bit) even or odd • Even • Odd • Good for checking single-bit errors 4-bit Example 7-bit Example 1 0 0 1

  41. Binary Logic • Operators • NOT If ‘x’ = 0 then NOT ‘x’ = 1 If ‘x’ = 1 then NOT ‘x’ = 0 • AND If ‘x’ = 1 AND ‘y’ = 1 then ‘z’ = 1 Otherwise ‘z’ = 0 • OR If ‘x’ = 1 OR ‘y’ = 1 then ‘z’ = 1 Otherwise ‘z’ = 0

  42. z = x = x’ Binary Logic • Truth Tables, Boolean Expressions, and Logic Gates AND OR NOT z = x • y = x y z = x + y

  43. Example of binary signals Logic Signals • Binary ‘0’ is representedby a “low” voltage(range of voltages) • Binary ‘1’ is representedby a “high” voltage(range of voltages) • The “voltage ranges” guardagainst noise

  44. Switching Circuits AND OR

  45. Homework • Mano • Chapter 1 • 1-2 • 1-7 • 1-9 • 1-10 • 1-11 • 1-16 • 1-18 • 1-20 • 1-24(a) • 1-29 • Write your family name in ASCII with odd parity • Decode the following ASCII string (with MSB = parity): • 110000110110111111101101111100001100000001010000010100110101010111010100 • Is the parity even or odd?

  46. Homework • Mano

  47. Homework

  48. Homework

  49. Homework

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